Prove $\operatorname{Var}(X+Y) \le 2\operatorname{Var}(X) + 2\operatorname{Var}(Y)$ where $X$ and $Y$ are not necessarily independent

Question

I know how to show that $\operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X,Y)$

I was also given the hint that I should use the triangle inequality to get

$|\operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X,Y)| \le |\operatorname{Var}(x)| + |\operatorname{Var}(Y)| + |2\operatorname{Cov}(X,Y)|$

Honestly, I have no idea where to go from here.

Answer 1

The difference is $$2E((X-\mu_X)^2)+2E((Y-\mu_Y)^2)-E((X+Y-\mu_X-\mu_Y)^2) =2E(X'^2)+2E(Y'^2)-E((X'+Y')^2)$$ where $X'=X-\mu_X$, $Y'=Y-\mu_Y$. Can you simplify that to something conveniently positive?

Answer 2

The key is to show that $|2 \operatorname{Cov}(X,Y)| \leq \operatorname{Var}(X) + \operatorname{Var}(Y)$.

Here's a hint towards that: Write $\hat{X} = X - \mu_X$ and $\hat{Y} = Y - \mu_Y$. Then note that $$E[(\hat{X} - \hat{Y})^2] \geq 0\,.$$

What happens if you expand the expectation?

Answer 3

Let $X'=X-EX,Y'=Y-EY$. Then the inequality to be proved is $E(X'+Y')^{2} \leq 2EX'^{2}+2EY'^{2}$ whic follows from the fact that $(a+b)^{2} \leq 2a^{2} +2b^{2}$ for any two real numbers $a$ and $b$.

Answer 4

Start as you did with: $$ Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y). $$ The Cauchy–Schwarz inequality then gives us: $$ 2Cov(X, Y) \leq 2[Var(X)]^{1/2}[Var(Y)]^{1/2}. $$ Finally, Young's inequality for products gives us: $$ 2[Var(X)]^{1/2}[Var(Y)]^{1/2} \leq Var(X) + Var(Y), $$ so we get, as desired: $$ Var(X + Y) \leq 2Var(X) + 2Var(Y). $$

Answer 5

If you know how to show $$\operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X,Y)\tag1$$ then you can show $$ \operatorname{Var}(X-Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) - 2\operatorname{Cov}(X,Y).\tag2 $$ Now add equations (1) and (2) to get $$ \operatorname{Var}(X+Y)+\operatorname{Var}(X-Y) = 2\operatorname{Var}(X) + 2\operatorname{Var}(Y). $$ Finally the fact that variance of anything is non-negative implies that $$ \operatorname{Var}(X+Y)\le \operatorname{Var}(X+Y)+\operatorname{Var}(X-Y). $$